\(\lim\limits_{x\rightarrow2}\dfrac{\sqrt{x+2}-2}{x^2-4}=\lim\limits_{x\rightarrow2}\dfrac{x-2}{\left(x-2\right)\left(x+2\right)\left(\sqrt{x+2}+2\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{1}{\left(x+2\right)\left(\sqrt{x+2}+2\right)}=\dfrac{1}{4.4}=\dfrac{1}{16}\)
a) lim ( \(\sqrt{x^2-x+1}-\sqrt{x^2+x+1}\)
x-> +∞
b) lim \(\dfrac{\sqrt{4x+1}-3}{x^2-4}\)
x-> 2
c) lim \(\dfrac{\sqrt{2x+5}-1}{x^2-4}\)
x-> -2
a) lim \(\dfrac{x\sqrt{x^2+1}-2x+1}{^3\sqrt{2x^3-2}+1}\)
x-> -∞
b) lim \(\dfrac{\left(2x+1\right)^3\left(x+2\right)^4}{\left(3-2x\right)^7}\)
x-> -∞
c) lim \(\dfrac{\sqrt{4x^2+x}+^3\sqrt{8x^3+x-1}}{^4\sqrt{x^4+3}}\)
x-> +∞
a) lim \(\dfrac{2x-\sqrt{3x^2+2}}{5x+\sqrt{x^2+2}}\)
x-> +∞
b) lim \(\sqrt{\dfrac{x^2+1}{2x^4+x^2-3}}\)
x-> ∞
cho \(\lim\limits_{x\rightarrow-\infty}\dfrac{a\sqrt{x^2+1}+2017}{x+2018}=\dfrac{1}{2}\); \(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+bx+1}-x\right)=2\). Tính P=4a+b
tính \(\lim\limits_{x\rightarrow0}\left(\dfrac{x}{\sqrt[7]{x+1}.\sqrt{x+4}-2}\right)\)
Tìm giơi han:
a) lim (x-> \(+\infty\)) \(\dfrac{\sqrt{x^2+1}+x}{5-2x}\)
b) lim (x->4) \(\left(\dfrac{\sqrt{15x+4}-\sqrt{x-3}-3}{-x+4}\right)\)
sorry, e k bt nhâp lim ..
a) lim \(\dfrac{3x^4-2x^5}{5x^4+x+4}\)
x-> -∞
b) lim \(\dfrac{x-1}{\sqrt{x^2-1}}\)
x-> +∞
lim (2+x) . \(\sqrt{\dfrac{x-1}{x^4+x^2+1}}\)
x-> + ∞
lim x\(\sqrt{\dfrac{x^2+1}{2x^4+x^2-3}}\)
x-> +∞
